Can you find area of the Yellow shaded region? | (Rectangle) | #math #maths | #geometry

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Can you find area of the Yellow shaded region? | (Rectangle) | #math #maths | #geometry

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[sergioaiex]

Nice Question @premath, thanks

(√5)² + (2√5)² = CE²
5 + 20 = CE²
CE = 5

a² + b² = (√5)²
a² + b² = 5

(5 - a)² + b² = (2√5)²
25 - 10a + (a² + b²) = 20
25 - 10a + 5 = 20
30 - 10a = 20
10a = 10
*a = 1*
*b = 2*

4/2 = c/4
*c = 8*

YSR = ½ (10 + 8) . 1


*YSR = 9 Square Units*

[wackojacko]

@ 7:27 one could follow the Lemmings and take the plunge or use Euclid's Theorem to save one's self. Either way outcome 100% survival cuz Lemmings do swim. 🙂

[prossvay]

∆ ABE is right triangle
CE^2=BC^2+BE^2
So CE=5
(BC)(BE)=(BD)(CE)
(2√5)(√5)=(BD)(5)
So BD=2
DE^2=BE^2-BD^2
DE^2=(√5)^2-(2^2)=1
So DE=1
CD=CE-DE=5-1=4
∆ADC~∆CDB
BD/CD==CD/AD
2/4=4/AD
So AD=8
So yellow shaded square units.❤❤❤

[juanalfaro]

EBC is a 1-2-sqrt (5) triangle. BCE=a, BEC=b --> CBD=b, DBE=a, ACD=b, CAD=a. Now CE=5 by Pythagoras, ED=1, BD=2, CD=4, AD=8. Then [ABEF] = [ADEF] + [DBE] = 8*1 + 1*2/2 = 8+1 = 9

[santiagoarosam]

Los tres triángulos rectángulos de la figura son semejantes---> Triángulo amarillo: 1²+2²=5---> Triángulo superior derecho: superior izquierdo: cateto largo =2*4=8---> Área amarilla =(8*1)+(1*2/2) =8+1=9 ud².
Gracias y saludos.

[marcelowanderleycorreia]

Very good, as always. Congratiulations professor!

[robertlynch]

I'm probably never going to get any traction on this but ...

The MOST overlooked theorem is H = AB/C ... which is to say, for a right triangle, the height of a line from intersection of hypotenuse to the right-angle apex is AB/C where A and B are the two legs and C is the hypotenuse. It ALWAYS works.

There is a similar theorem for the "two parts of the divided hypotenuse". The side close to [A] has formula AA/C and the side close to B has BB/C

Using just these then gives all the measures, leading the area to be (1 * 8) + 1/2( 2 * 8 ) = 9 units^2

[phungpham]

1/ Note that all the right triangles are similar and the ratio of their relevant short leg/ long leg= 1/2.
2/ Focus on the triangle EDB. Label ED= a —>DB= 2a —> sqa+ sq(2a)=sqEB=5
—> a= 1 DB=2—>DC=4 and AD=8
Area of the yellow ABEF = Area of the rectangle ADEF+ Area of the triangle EDB= 8+1= 9 sq units😅

[michaelstahl]

Great video again. Thank`s Sire . I used Euclid`s theorem and I got the same result before watching your video .

[marcgriselhubert]

*In triangle EBC: EC^2 = BC^2+BE^2 = 20 + 5 = 25,
so EC = 5
*Triangles BAC and ECB are similar (equal angles),
so BA/EC = BC/EB or BA/5 = (2.sqrt(5))/sqrt(5) = 2
so we have BA = 10.
*In triangle ABC: AC^2 = BA^2 - BC^2 = 100 -20 = 80,
so AC = sqrt(80) = 4.sqrt(5)
*Area of ABC = AC.BC = (4.sqrt(5)).(2.sqrt(5)) = 40
Area of ABC is also AB.CD = 10.CD, so CD = 4
Then DE = CE - DE = 5 - 4 = 1
*Area of DBC = DB.EC = DB.5
Area of DBC is also BC.EB = (2.sqrt(5)).(sqrt(5)) = 10
So DB = 2 and FE = AD = AB -DB = 10 - 2 = 8
*Area of trapezoïd AFEB = ((AB + FE)/2).DE
= ((10 + 8)/2).1 = 9.

[gaylespencer]

Followed those methods for most of the way. Then used a little trig. to get the length of AD.
Tan (63.43494...) times 4 = 8.

[yalchingedikgedik]

Thanks prof.
Thanks PreMath
Very useful method
Many likes ❤❤❤❤❤
Good luck with glades

[murdock]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = AD + BD = n + m; CD = h; BC = 2√5; BE = √5
sin⁡(EBC) = sin⁡(BCA) = sin⁡(ADC) = 1; ∎ADEF → AD = EF = n; AF = DE = k → CE = 5
BCD = EBD = CAD = δ → sin⁡(δ) = √5/5 → cos⁡(δ) = 2√5/5 → tan⁡(δ) = 1/2 →
m = (2√5)(√5)/5 = 2 → k = 1 → h = 4 → n = 8 → green shaded area = k(n + m/2) = 9

[rabotaakk-nwnm]

7:45 [yellow]=[ABEF] (right trapezoid)
AB=8+2=10; EF=AD=8; AF=DE=1
sq.un.

[quigonkenny]

As ∠BDE = ∠EBC = 90° and ∠CEB is common, ∆BDE and ∆EBC are similar triangles. As ∠BDE = ∠BCA = 90° and ∠CAB = ∠EBD as opposite angles, ∆BCA is also similar to ∆BDE.

Triangle ∆BDE:
DE/BD = EB/BC
DE/BD = √5/2√5 = 1/2
BD = 2DE

Let DE = k.

BD² + DE² = EB²
(2k)² + k² = (√5)²
4k² + k² = 5
5k² = 5
k² = 1 ==> k = 1

DE = 1, BD = 2

Triangle ∆BCA:
BC/AB = DE/EB
2√5/AB = 1/√5
AB = 2√5(√5) = 10

Trapezoid FEBA:
Aₜ = h(a+b)/2 = DE(FE+AB)/2
Aₜ = 1((10-2)+10)/2
Aₜ = (8+10)/2 = 9 sq units

[shahriarhossain]

We can also use similarity of the triangles to solve

[misterenter-izrz]

A bit reluctant to do, for clumsy computation, h=10/5=2, 5 divided in to 4, 1, then the area is (8+10)×1/2=9.😅

[davidstecchi]

Once you solved or all the side lengths, why didn' you ust use the ormula for the area o a trapezoid instead o combining a square and a rectangle?

[unknownidentity]

Let's find the area:
.
..
...
....


The triangle BCE is a right triangle, so we can apply the Pythagorean theorem:

CE² = BC² + BE² = (2√5)² + (√5)² = 20 + 5 = 25 ⇒ CE = 5

Now let's have a look at the interior angles of some triangles:

BCE: ∠CBE = 90° ∠BCE = α ∠BEC = β
BCD: ∠BDC = 90° ∠BCD = α ∠CBD = β
ABC: ∠ACB = 90° ∠BAC = α ∠ABC = β
BDE: ∠BDE = 90° ∠DBE = α ∠BED = β

Therefore all the triangles BCE, BCD, ABC and BDE are similar and we can conclude:

AB/BC = CE/BE
AB/(2√5) = 5/√5 ⇒ AB = (2√5)*(5/√5) = 10

CD/BC = BC/CE
CD/(2√5) = 2√5/5 ⇒ CD = (2√5)*(2√5/5) = 4

BD/BE = BC/CE
BD/√5 = 2√5/5 ⇒ BD = √5*(2√5/5) = 2

DE = CE − CD = 5 − 4 = 1
EF = AD = AB − BD = 10 − 2 = 8

Now we are able to calculate the area of the yellow trapezoid:

A(ABEF) = (1/2)*(AB + EF)*DE = (1/2)*(10 + 8)*1 = 9

Best regards from Germany